Question: $ \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-\sqrt{9-3y^2}}^{\sqrt{9-3y^2}} \int_{-3}^3 dx \, dy$ (Choice B) B $ \int_{-3}^3 \int_{-\sqrt{9-3y^2}}^{\sqrt{9-3y^2}} dx \, dy$ (Choice C) C $ \int_{-3}^3 \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} dx \, dy$ (Choice D) D $ \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \int_{-3}^3 dx \, dy$
Solution: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $-3 < x < 3$ and $-\sqrt{9 - x^2} < y < \sqrt{9 - x^2}$. Therefore: ${2}$ ${4}$ ${\llap{-}2}$ ${\llap{-}4}$ ${2}$ ${4}$ ${\llap{-}2}$ ${\llap{-}4}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $-3 < y < 3$. Now we can define $x$ in terms of $y$. $-\sqrt{9 - y^2} < x < \sqrt{9 - y^2}$. We want to pay attention especially to how this $x$ bound works at the edge of the $y$ bound. For example, at $y = -3$, the $x$ bound makes $x = 0$ as expected. In conclusion, the double integral after switching bounds is: $ \int_{-3}^3 \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} dx \, dy$